J_n = \int \sin^n x \; dx = \int \sin^{n-1} x \cdot \sin x\; dx = -\cos x \cdot \sin^{n-1} x + \int (n-1) \sin^{n-2} x \cdot \cos^2 x \; dx =
= -\cos x \cdot \sin^{n-1} x + (n-1) \int \sin^{n-2} x \cdot (1 - \sin^2 x) \; dx = -\cos x \cdot \sin^{n-1} x + (n-1) \int \sin^{n-2} x \; dx - (n-1) \int \sin^n x \; dx
więc
n J_n = -\cos x \cdot \sin^{n-1} x + (n-1) J_{n-2}
więc
\displaystyle J_n = - \frac 1 n \cos x \cdot \sin^{n-1} x + \frac {n-1} n J_{n-2}
\displaystyle L_0 = \int_0^{\pi/2} 1 \; dx = \frac {\pi} 2
\displaystyle L_1 = \int_0^{\pi/2} \sin x \; dx = - \cos x |_0^{\pi/2} = 1
\displaystyle L_{2n} = -\frac 1 {2n} \sin^{2n-1} x \cos x |_0^{\pi/2} + \frac {2n-1} {2n} L_{2n-2} = \frac {2n-1} {2n} L_{2n-2} = \frac {(2n-1)(2n-3)} {(2n)(2n-2)} L_{2n-4} = \cdots = \frac {(2n-1)(2n-3)\cdots 3 \cdot 1} {(2n)(2n-2) \cdots 4 \cdot 2} L_0 = \frac {(2n-1)!!} {(2n)!!} \cdot \frac {\pi} 2
L_{2n+1} podobnie.
\displaystyle \int_0^{\infty} \frac {dx} {(1+x^2)^n} = M_n \bigg |_0^{\infty} = \frac 1 {2n-2} \frac x {(1+x^2)^{n-1}} \bigg |_0^{\infty} + \frac {2n-3} {2n-2} M_{n-1} \bigg |_0^{\infty} = \frac {2n-3} {2n-2} M_{n-1} \bigg |_0^{\infty} = \frac {(2n-3)(2n-5)} {(2n-2)(2n-4)} M_{n-2} \bigg |_0^{\infty} = \cdots = \frac {(2n-3)!!} {(2n-2)!!} M_1 = \frac {(2n-3)!!} {(2n-2)!!} \frac {\pi} 2
Powołujemy się na fakt, że \displaystyle \int_{-\infty}^{\infty} e^{-x^2} \; dx = \sqrt {\pi} (http://en.wikipedia.org/wiki/Gaussian_integral).
\displaystyle \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-\frac 1 2 x^2} \; dx
Podstawiamy y = \frac 1 {\sqrt 2} x. \sqrt 2 dy = dx.
\displaystyle \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-y^2} \; dy \cdot \sqrt 2 = \frac {\sqrt {2\pi}} {\sqrt {2\pi}} = 1
Z drugiej strony, to jest szczególny przypadek całki z zadania 8.
Szczególny przypadek całki z zadania 9.
Szczególny przypadek całki z zadania 10.
\displaystyle \frac 1 {\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} exp\{-\frac {(x-\mu)^2} {2\sigma^2}\} \; dx
Podstawiamy t = \frac {x - \mu} {\sqrt 2 \sigma}. dt \sqrt 2 \sigma = dx.
\displaystyle \frac 1 {\sqrt{2\pi} \sigma} \int_{-\infty}^{\infty} exp\{-t^2\} \sqrt 2 \sigma \; dt = \frac {\sqrt {2 \pi} \sigma}{\sqrt {2 \pi} \sigma} = 1
Podstawmy t = \frac {x - \mu} {\sigma}; x = \sigma t + \mu; dx = \sigma dt.
\displaystyle \Big(-\exp\{-\frac {t^2} {2}\}\Big)' = -\exp\{-\frac {t^2} {2}\} \cdot (-t) = t\exp\{-\frac {t^2} {2}\}
A więc
\displaystyle \frac 1 {\sqrt {2 \pi} \sigma} \int_{-\infty}^{\infty} x \exp \{ - \frac {(x-\mu)^2} {2 \sigma^2} \} \; dx = \frac 1 {\sqrt {2 \pi}} \int_{-\infty}^{\infty} \sigma t \exp\{ - \frac {t^2} 2 \} \; dt + \frac 1 {\sqrt {2 \pi}} \int_{-\infty}^{\infty} \mu \exp\{ - \frac {t^2} 2 \} \; dt = - \frac {\sigma} {\sqrt {2 \pi}} \exp\{-\frac {t^2} {2}\} \Big |_{-\infty}^{\infty} + \frac {\sqrt{2 \pi} \mu} {\sqrt{2 \pi}} = 0 + \mu = \mu
Podstawmy t = \frac {x - \mu} {\sigma}; x = \sigma t + \mu; dx = \sigma dt.
\displaystyle \frac 1 {\sqrt {2 \pi} \sigma} \int_{-\infty}^{\infty} (x-\mu)^2 \exp \{ - \frac {(x-\mu)^2} {2 \sigma^2} \} \; dx = \displaystyle \frac {\sigma} {\sqrt {2 \pi} \sigma} \int_{-\infty}^{\infty} (\sigma t)^2 \exp \{ - \frac {t^2} {2} \} \; dt = \displaystyle \frac {\sigma^2} {\sqrt {2 \pi}} \int_{-\infty}^{\infty} -t \cdot (-t \exp \{ - \frac {t^2} {2} \}) \; dt = -\displaystyle \frac {\sigma^2} {\sqrt {2 \pi}} t \exp \{ - \frac {t^2} {2} \} \big |_{-\infty}^{\infty} + \displaystyle \frac {\sigma^2} {\sqrt {2 \pi}} \int_{-\infty}^{\infty} \exp \{ - \frac {t^2} {2} \} \; dt = 0 + \frac {\sigma^2 \sqrt{2 \pi} } {\sqrt{2 \pi}} = \sigma^2